Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

C 1 = 5μF, C 2 = 10 μF, C 3 = 2 μF Calculate the equi. Capacitance. a)6 μF b)2 μF c)4 μF d)8 μF

a. Find the equivalent capacitance between points a and b. b. What fraction of the total energy is stored in the 7 µF capacitor, if the circuit has been charged by a 12 volt battery? a µF b.46%

C µF, q = 28.8 µC C 2  C 24 = 12 µF C 1234 = 3 µF q =36 µC Δq = 7.2 µC C 1 = C 3 = 8.00 μF, C 2 = C 4 = 6.00 μF, V = 12V When the switch S is closed, how much charge flows through point P

Question A parallel-plate capacitor has a plate area of 0.3m 2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10 -6 C then the force exerted by one plate on the other has a magnitude of about: A. 0B. 5NC. 9ND. 1 x10 4 N E. 9 x 10 5 N

Question A parallel-plate capacitor has a plate area of 0.3m 2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10 -6 C then the force exerted by one plate on the other has a magnitude of about: A.0 B.5N C.0. 9N D.1 x10 4 N E.9 x 10 5 N The electric field = σ/2ε o why?

q-q q'q' q q -q' -q V V V V'V' ( )

q -q q'q' q q -q' -q V V V V'V' ( )

If the areas are A 1 and A-A 1. C µF, q = 28.8 µC C 2  C 24 = 12 µF C 1234 = 3 µF q =36 µC Effect of a dielectric : C  κC

The force on a filling dielectric as it is inserted between the parallel plates of a capacitor. x L With the battery connected, U 1 = ½CV 2 With the battery disconnected, U 2 = Q 2 /2C With the battery connected, since x is increasing downwards, a negative force is upwards, pushing the dielectric away. With the battery disconnected, the force is positive and pointed downwards, pulling in the dielectric. The force is proportional to (κ-1) and inversely to L. To make the argument simple, we have omitted the fact that the field at the end includes the fringe field which depends on distance between the plates. The result including fringe fields is quantitatively different. See S. Margulies, American Journal of Physics, V. 52, p 515 (1984)

A question What is the equivalent capacitance between the points A and B? A.1 μF B. 2 μF C.4 μF D.10μF E.None of these AB What would a 10V battery do, i.e. how much charge will it provide, when it is connected across A and B? 40 μC (20 μC) on 2 μF

HITT A parallel-plate capacitor has a plate area of 0.2m 2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 10 6 V/m between the plates, the magnitude of the charge on each plate should be: A. 8.9 x CB. 1.8 x C C. 3.5 x CD. 7.1 x C E. 1.4 x C

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All C’s are 8.00 nF. The battery is 12 V. What is the equivalent capacitance? C 12 = 4 nF C 123 = 12 nF Q 123 = C 123 x V = 144 nC Q 3 = C 3 x V = 96 nC Q 12 = C 12 x V = 48 nC U 123 = ½ C 123 V 2 = ½ x 12x10 -9 x12 2 = 864 nJ U 1 = ½ C 1 V 1 2 = ½ x 8x10 -9 x6 2 = 144 nJ = U 2 U 3 = ½ x 8x10 -9 x12 2 = 576 nF C 3 stores most energy, also the highest electric field and most charge, the most stressed part of the circuit.

Series combinations reduce the capacitance. Equal C reduce by the number involved. In parallel the capacitance increases.  A basket of 4 capacitors, each of C = 6 nF. How can you arrange them to get a)1.5 nF b)2 nF g) 2.4 nF c)3 nFh) 3.6 nF d)4 nFi) 4.5 nFj) 6 nF e)12 nFk) 18 nF f)24 nF

clicker All C’s are 8.00 nF. The battery is 12 V. What is the equivalent capacitance? a.4 nF b.6 nF c.8 nF d.10 nF e.12 nF

Circuits All capacitors being the same, rank the equivalent capacitances of the four circuits.

summary Capacitance Parallel plates, coaxial cables, Earth Series and parallel combinations Energy in a capacitor Dielectrics Dielectric strength

A question Each of the four capacitors shown is 500 μ F. The voltmeter reads 1000V. The magnitude of the charge, in coulombs, on each capacitor plate is: A. 0.2B. 0.5C. 20D. 50E. none of these